Problem: Simplify and expand the following expression: $ \dfrac{t - 6}{t - 3}-\dfrac{t}{3t - 7} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(t - 3)(3t - 7)$ Multiply the first term by $\dfrac{3t - 7}{3t - 7}$ $ \begin{align*} \dfrac{t - 6}{t - 3} \times \dfrac{3t - 7}{3t - 7} & = \dfrac{(t - 6)(3t - 7)}{(t - 3)(3t - 7)} \\ & = \dfrac{3t^2 - 25t + 42}{(t - 3)(3t - 7)}\end{align*} $ Multiply the second term by $\dfrac{t - 3}{t - 3}$ $ \begin{align*} \dfrac{t}{3t - 7} \times \dfrac{t - 3}{t - 3} & = \dfrac{(t)(t - 3)}{(3t - 7)(t - 3)} \\ & = \dfrac{t^2 - 3t}{(3t - 7)(t - 3)}\end{align*} $ Now we have: $ = \dfrac{3t^2 - 25t + 42}{(t - 3)(3t - 7)} - \dfrac{t^2 - 3t}{(3t - 7)(t - 3)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{3t^2 - 25t + 42 - (t^2 - 3t)}{(t - 3)(3t - 7)} $ $ = \dfrac{3t^2 - 25t + 42 - t^2 + 3t}{(t - 3)(3t - 7)} $ $ = \dfrac{2t^2 - 22t + 42}{(t - 3)(3t - 7)}$ Expand the denominator: $ = \dfrac{2t^2 - 22t + 42}{3t^2 - 16t + 21}$